Answer:
The volume of gas in an early diving bell full of air at sea level is halved at 10 m according to Boyle’s law
Explanation:
;at 20 m pressure is 300 kPa absolute and the gas is compressed into one third the volume.
HOPE THIS HELPS i did this before
Derive the equation relating the total charge Q that flows through a search coil (Conceptual Example 29.3) to the magnetic-field magnitude B. The search coil has N turns, each with area A, and the flux through the coil is decreased from its initial maximum value to zero in a time Δt. The resistance of the coil is R, and the total charge is Q=IΔt, where I is the average current induced by the change in flux.
Answer:
Q= NBA/R
Explanation:
Check attachment for derivation
The equation relating the total charge, magnitude, turns, time will be "[tex]\frac{NBA}{R}[/tex]".
Magnetic fieldAccording to the question,
Resistance = R
Total charge = Q
Current = I
Number of turns = N
Time = Δt
and,
Q = IΔt ...(equation 1)
We know the flux,
→ [tex]\Phi[/tex] = NBA
Emf induced,
ε = [tex]\frac{- \Delta \Phi}{\Delta t}[/tex]
Δ[tex]\Phi[/tex] = [tex]\Phi_2 - \Phi_1[/tex]
then,
ε = [tex]\frac{NBA}{\Delta t}[/tex]
As we know, Voltage (V) = iR
then, ε = [tex]\frac{NBA}{\Delta t}[/tex] = iR
i = [tex]\frac{NBA}{R \Delta t}[/tex]
Hence, by applying the values in "equation 1"
→ Q = iΔt
= [tex]\frac{NBA}{R \Delta t}[/tex] × Δt
= [tex]\frac{NBA}{R}[/tex]
Thus the response above is correct.
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Modified Newtonian dynamics(MoND)proposes that, for small accelerations, Newton’s second law, F = ma, approaches the form F = ma2/a0, where a0 is a constant.
(a) (10 points) Show how such a modified version of Newton’s second law can lead to flat rotation curves, without the need for dark matter.
(b) (10 points) Alternatively, propose a new law of gravitation to replace F = GMm/r2 at distances greater than some characteristic scale r0 so that again, you can explain the observed flat rotation curved of galaxies without dark matter.
Answer:
Explanation:
The two pictures attached here shows the solution to the two questions from the problem. thank you and I hope it helps you
An athlete is working out in the weight room. He steadily holds 50 kilograms above his head for 10 seconds. Which statement is true about this situation?
Answer:
Answer: the true statement form the given statements is “the athletes is not doing any work because he does not move weight”
Explanation:
The athlete isn’t doing any work because he doesn’t move the weight is the correct statement.
What is Work? Work is the energy transferred to or from an object via the application of force along a displacement.Work = Force x Displacement.How to solve this Problem?The weight of an object given is 50kgsThe time of holding an object given is 10 secondsWe need to justify the statements
Here ,
There is no displacement that means displacement is zero.If displacement is zero then work done will also be zeroHence there is no work done by the athlete
Therefore ,The athlete isn’t doing any work because he doesn’t move the weight is the correct statement
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The universal law of gravitation states that the force of attraction between two objects depends on which quantities?
the masses of the objects and their densities
the distance between the objects and their shapes
the densities of the objects and their shapes
the masses of the objects and the distance between them
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Subm
Kandretum
Answer:depends on the masses of the objects and the distance between them
Explanation:
According to Newton's law of universal gravitation,the force of attraction between two objects depends on the masses of the objects and the distance between them
Is mercury (the planet) rocky or gaseous(meaning relating to or having the characteristics of a gas.)
Answer:
Mercury is rocky
Explanation:
Answer:
Rocky
Explanation:
It has no atmosphere so it cannot hold gas.
Two wires, both with current out of the page, are next to one another. The wire on the left has a current of 1 A and the wire on the right has a current of 2 A. We can say that:
A. The left wire attracts the right wire and exerts twice the force as the right wire does.
B. The left wire attracts the right wire and exerts half the force as the right wire does.
C. The left wire attracts the right wire and exerts as much force as the right wire does.
D. The left wire repels the right wire and exerts twice the force as the right wire does.
E. The left wire repels the right wire and exerts half the force as the right wire does.
F. The left wire repels the right wire and exerts as much force as the right wire does.
Answer:
C. The left wire attracts the right wire and exerts as much force as the right wire does.
Explanation:
To know what is the answer you first take into account the magnetic field generated by each current, for a distance of d:
[tex]B_1=\frac{\mu_oI_1}{2\pi d}=\frac{\mu_o}{2\pi d}(1A)\\\\B_2=\frac{\mu_oI_2}{2\pi d}=\frac{\mu_o}{2\pi d}(2A)=2B_1\\\\[/tex]
Next, you use the formula for the magnetic force produced by the wires:
[tex]\vec{F_B}=I\vec{L}\ X \vec{B}[/tex]
if the direction of the L vector is in +k direction, the first wire produced a magnetic field with direction +y, that is, +j and the second wire produced magnetic field with direction -y, that is, -j (this because the direction of the magnetic field is obtained by suing the right hand rule). Hence, the direction of the magnetic force on each wire, produced by the other one is:
[tex]\vec{F_{B1}}=I_1L\hat{k}\ X\ B_2(-\hat{j})=I_1LB_2\hat{i}=(2A^2)\frac{L\mu_o}{2\pi d}\hat{i}\\\\\vec{F_{B2}}=I_2L\hat{k}\ X\ B_2(\hat{j})=I_2LB_1\hat{i}=-(2A^2)\frac{L\mu_o}{2\pi d}\hat{i}[/tex]
Hence, due to this result you have that:
C. The left wire attracts the right wire and exerts as much force as the right wire does.
Describe at least two different objects that you think are magnetic and why do you think this objects are magnetic
Answer:
Magnetic letters Screws and bolts
Explanation:Magnetic letters have to be magnetic so they can stick to things like walls and fridges. Screws and bolts are magnetic becaus they become easier to to screw onto things since the tool that you use to screw it in is also magnetic.
Answer:
File cabins and nails/screw,
Explanation:
Because they are ferromagnetic this means that they can be magnetized. Magnetic objects are made of metal and metals do attract magnet.
What happens if you move a magnet near a could of wire
Answer:
The wire would stick to the magnet????????????????????????
Explanation:
Air is matter which backs best support the statement
Answer: A. Balloons can be filled with air.
C. Air has mass.
Explanation:
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Balloons are able to be filled with air and air has mass.
In this circuit the battery provides 3 V, the resistance R1 is 7 Ω, and R2 is 5 Ω. What is the current through resistor R2? Give your answer in units of Amps. An Amp is 1 Coulomb of charge flowing through a cross-sectional area of the wire per second - that's a lot of charge per second and will warm up a typical wire quite a bit! Most devices have circuits with larger resistors - kLaTeX: \OmegaΩ (103 LaTeX: \OmegaΩ) and MLaTeX: \OmegaΩ (106 LaTeX: \OmegaΩ) are common.
Answer:
The current pass the [tex]R_2[/tex] is [tex]I = 0.25 A[/tex]
Explanation:
The diagram for this question is shown on the first uploaded image
From the question we are told that
The voltage is [tex]V = 3V[/tex]
The first resistance is [tex]R_1 = 7 \Omega[/tex]
The second resistance is [tex]R_2 = 5 \Omega[/tex]
Since the resistors are connected in series their equivalent resistance is
[tex]R_{eq} = R_1 +R_2[/tex]
Substituting values
[tex]R_{eq} = 7 + 5[/tex]
[tex]R_{eq} = 12 \Omega[/tex]
Since the resistance are connected in serie the current passing through the circuit is the same current passing through [tex]R_2[/tex] which is mathematically evaluated as
[tex]I = \frac{V}{R_{eq}}[/tex]
Substituting values
[tex]I = \frac{3}{12}[/tex]
[tex]I = 0.25 A[/tex]
In order to get going fast, eagles will use a technique called stooping, in which they dive nearly straight down and tuck in their wings to reduce their surface area. While stooping, a 6- kg golden eagle can reach speeds of up to 53 m/s . While golden eagles are not very vocal, they sometimes make a weak, high-pitched sound. Suppose that while traveling at maximum speed, a golden eagle heads directly towards a pigeon while emitting a sound at 1.1 kHz. The emitted sound has a sound intensity level of 30 dB when heard at a distance of 5 m .A) Model this stooping golden eagle as an object moving at terminal velocity. The eagle’s drag coefficient is 0.5 and the density of air is 1.2 kg/m 3 . What is the effective cross-sectional area of the eagle’s body while stooping?B) What is the doppler-shifted frequency that the pigeon will hear coming from the eagle?C) Consider the moment when the pigeon is 5 m away from the eagle. At the pigeon’s position, what is the intensity (in W/m^2 ) of the sound the eagle makes?D) The golden eagle slams into the 250- g pigeon, which is initially moving at 10 m/s in the opposite direction (toward the eagle). The eagle grabs the pigeon in its talons, and they move off together in a perfectly inelastic collision. How fast do they move after the collision?
Answer:
Check the explanation
Explanation:
Part A
F = CA
this drag force balances the weight = 6X 9.8
so
6X9.8 = 0.5 X A X0.5 X 1.2 X 532
A= 0.069 m2
Part B
here the sorce is moving and the observer is at rest
so f= f(- 1 - 1
f = 1.1X10 343 343 – 53
f' = 1.3 KHz
Part C:
given the intensity = 30 dB
we know that I dB = 10 log (I(W/m2))
so we get I (W/m2) = 1000
Part D : The catch
Given that U1 = 53 M1 = 6 kg
U2 =-10 M2=0.25
V1=V2
now conserving momentum
6 X 53 -0.25 X10 =(6+0.25)V
V= 50.48 m/sec
I need some help!!!!!!!!!
Answer:
The Object will immediately begin moving toward the left
Explanation:
Because the force of thirteen is greater than ten and applied to the opposite side
1. You are playing with a jump rope that is tied at both ends. You untie one end, hold it taut and wiggle the end up and down sinusoidally with frequency 2.00Hz and amplitude 0.075m. At time t=0, the end has a maximum positive displacement and is instantaneously at rest. Assume no wave bounces back from the far end to change the pattern. What is the equation for the displacement of the wave? What is the displacement at a point 3.00m from the end .
Answer:
[tex]f(x=3.00m)=0.075mcos(\frac{2\pi(2.00Hz)}{v}(3.00m))[/tex]
Explanation:
To find the equation of the wave you use the general equation for a wave, given by:
[tex]f(x)=Acos(k x-\omega t)[/tex]
A: amplitude of the wave = 0.075m
k: wave number
you select a cosine function because for x=0 and t= 0 you get a maximum displacement.
To find the displacement of the wave for x=0 you can consider that the form of the wave is independent of time t.
Then, you calculate k:
[tex]k=\frac{\omega}{v}=\frac{2\pi f}{v}[/tex]
Thus, you need the value of the speed of the wave (you only have the frequency f), in order to calculate f(x), for x=3.00m:
[tex]f(x=3.00m)=0.075mcos(\frac{2\pi(2.00Hz)}{v}(3.00m))[/tex]
A pipe branches symmetrically into two legs of length L, and the whole system rotates with angular speed ω around its axis of symmetry. Each branch is inclined at angle α to the axis of rotation. Liquid enters the pipe steadily, with zero angular momentum, at volume flow rate Q. The pipe diameter, D, is much smaller than L. Obtain an expression for the external torque required to turn the pipe. What additional torque would be required to impart angular acceleration ω_ ?
Answer:
Check the explanation
Explanation:
Kindly check the attached images below to see the step by step explanation to the question above.
Which of the following BEST summarizes the relationship between groups and culture and critical thinking?
Answer:
Groups and culture helps in influencing our values,ethics and beliefs. This influence should always be questioned through the process of thinking critically.
This best summarizes the relationship between groups and culture and critical thinking.
which one of the following statements is true? A.in an elastic collision,only momentum is conserved B. in any collision,both momentum & kinetic energy are conserved C.in an inelastic collision,both momentum & kinetic energy are conserved D.in an elastic collision,only kinetic energy is conserved
Answer:
option C is correct
................
Answer:
C- in an inelastic collision, both momentum & kinetic energy are conserved
Explanation:
Took the test
A brick is dropped from a high scaffold. a. What is its velocity after 4.0s ?
b. How far does the brick fall during this time ?
Answer:
A: 1.962
B: 3.924
Explanation:
g = G *M /R^2
g = 9.807*M/R^2 the gravitational constant of ground level on earth is about 9.807
g = 9.807*5lbs/R^2 the average brick is about 5 pounds.
g = 9.807*5*10^2. I'm assuming the height is around ten feet to help you out.
with these numbers plugged in you get an acceleration of 0.4905 a final velocity after 4 seconds 1.962. It's height fallen after 4 seconds is 3.924.
( M = whatever the brick weighs it's not specified in the question)
(R = the distance from the ground or how high the scaffold is)
(hopefully you can just plug your numbers in there hope this helps)
A 50 Ohm resistance causes a current of 5 milliamps to flow through a circuit connected to a battery. What is the power in the circuit?
Answer:0.00125 watts
Explanation:
resistance=50 ohms
Current=5 milliamps
Current=5/1000 amps
Current=0.005 amps
Power=(current)^2 x (resistance)
Power=(0.005)^2 x 50
Power=0.000025 x 50
Power=0.00125 watts
Answer:
if a circuit is powered by 100v and has an overall resistance of 50 ohms 2 amps flows through the circuit
Explanation:
You are designing an optical fiber scope for directing light into a confined area. You want to keep light within the fiber. Based on the specifications, you know that the greatest angle that the light will make with the horizontal is no greater than 25⁰. Assuming you will be using the scope in the body which has the same index of refraction of water (n = 1.33). What is the minimum index of refraction n2 required for the design to be functional?
Answer:
Explanation:
For entry of light into tube of unknown refractive index
sin ( 90 - 25 ) / sinr = μ , μ is the refractive index of the tube , r is angle of refraction in the medium of tube
r = 90 - C where C is critical angle between μ and body medium in which tube will be inserted.
sin ( 90 - 25 ) / sin( 90 - C) = μ
sin65 / cos C = μ
sinC = 1.33 / μ , where 1.33 is the refractive index of body liquid.
From these equations
sin65 / cos C = 1.33 / sinC
TanC = 1.33 / sin65
TanC = 1.33 / .9063
TanC = 1.4675
C= 56°
sinC = 1.33 / μ
μ = 1.33 / sinC
= 1.33 / sin56
= 1.33 / .829
μ = 1.6 Ans
What is an independent variable?
A. A variable that is intentionally changed during an experiment
B. A variable that depends on the experimental variable
C. A variable that is not used in an experiment
D. A variable that is unknown during the experiment
Answer:
The answer is A
Explanation:
Independent variables don't have to depend on other factors of the experiment because they're independent
Answer:
A.
Explanation:
Independent variables don't have to depend on other factors of the experiment because they're independent.
A particular coil has 100 turns and a diameter of 6.0 m. When it's time for a measurement, a 4.5 A current is turned on. The large diameter of the coil means that the field in the water flowing directly above the center of the coil is approximately equal to the field in the center of the coil. The field is directed downward and the water is flowing east. The water is flowing above the center of the coil at 1.5 m/s .
What is the magnitude of the field at the center of the coil?
Answer:
The magnetic field at the center of the coil = 5.23 * 10 ^ -5 T
Explanation:
Information from the question:
Number of turns of the coil = 100 turns
The diameter of the coil = 6 m
The radius of the coil = diameter / 2 = 3 m
The coil current = 2.5 A
Formula : The Magnetic field at the center of the coil =
k * number of turns * current / 2 * radius
Therefore, The Magnetic field at the center of the coil=
(4 * [tex]\pi[/tex] * 10 ^ -7 * 100 * 2.5 ) / (2 * 3)
The Magnetic field at the center of the coil = 5.23 * 10 ^ -5 T
(20) A rocket is launched vertically. At time t = 0 seconds, the rocket’s engine shuts down. At the time, the rocket has reached an altitude of 500m and is rising at a velocity of 125 m/s. Gravity then takes over. The height of the rocket as a function of time is h(t)=-9.8/2 t^2+125t+500,t>0. Using your function file from HW2A: Generate a plot of height (vertical axis) vs. time (horizontal axis) from 0 to 30 seconds. Include proper axis labels. Find the maximum height and the time at which it occurs: Analytically, showing your steps and equations. This part should be done entirely in the write-up: no coding Using the data cursor on the plot. Using the MAX function on your data from part (a) Using FMINSEARCH on your m file Comment on the differences between the methods. How closely does each method match the "true" (analytical) value? Find the time when the rocket hits the ground: Analytically, showing your equations. This part should be done entirely in the write-up: no coding Using the data cursor on the plot. Using FZERO on your m file Comment on the differences between the methods in each of part (B) and (C). How closely does each method match the "true" (analytical) value? Use a quantitative comparison to make your argument.
Answer:
Explanation:
Given that,
h(t) = -9.8t² / 2 + 125t + 500
for t > 0.
At t = 0, the rocket is at height h = 500m, at a velocity of Vo = 125m/s.
We want to find the maximum height reached by rocket
Using mathematics maxima and minima
let find the turning point when dh/dt = 0
dh/dt = -9.8t + 125
dh / dt = 0 = -9.8t + 125
9.8t = 125
t = 125 / 9.8
t = 12.76s
Let find the turning point to know if this time t = 12.76 is maximum or minimum point
Let find d²h / dt²
d²h / dt² = -9.8
Since, d²h/dt² < 0, then, at t = 12.76s is the maximum points.
Then, the maximum height reached is
h = -9.8t² / 2 + 125t + 500
h = -9.8(12.76)² / 2 + 125(12.76) + 500
h = -797.80 + 1595 + 500
h = 1297.2 m
The maximum height reached is 1297.2 m
From the attachment, the maximum height is 1297.2m at t = 12.76sec.
Comment, the result are the same for both the analysis aspect and the graphical aspect.
A large boulder falls from an underwater ledge and crashes on the floor of the ocean. A pod of dolphins is 800 meters away. Determine how long it takes for the sound of the crash to reach the dolphins.
0.5 sec
0.3 sec
2.3 sec
1.9 sec
Answer:
0.5
Explanation:
Answer:
0.539
Explanation:
Why does current flow in a coil when a magnet is pushed in and out of the coil ?
Answer:
So the induced current opposes the motion that induced it (from Lenz's Law). When we pull the magnet out, the left hand end of the coil becomes a south pole (to try and hold the magnet back). Therefore the induced current must be flowing clockwise.
hope this helps u...
Find the frequency of the 4th harmonic waves on a violin string that is 48.0cm long with a mass of 0.300 grams
and is under a tension of 4.00N.
Answer:
The frequency of the 4th harmonic of the string is 481.13 Hz.
Explanation:
When a stretch string fixed at both ends is set into vibration, it produces its lowest sound of possible note called the fundamental frequency. Under certain conditions on the string, higher frequencies called harmonics or overtones can be produced.
The frequency of the forth harmonic is the third overtone of the string and can be determined by:
f = [tex]\frac{2}{L}[/tex][tex]\sqrt{\frac{T}{m} }[/tex]
Given that; L = 48.0 cm = 0.48 m,
m = 0.3 g = 0.0003 Kg,
T = 4.0 N,
f = [tex]\frac{2}{0.48}[/tex][tex]\sqrt{\frac{4}{0.0003} }[/tex]
f = 4.1667 × 115.4701
= 481.1252
f = 481.13 Hz
The frequency of the 4th harmonic of the string is 481.13 Hz.
A cobalt-60 source with activity 2.60×10-4 Ci is embedded in a tumor that has
mas 0.20 kg. The source emits gamma photons with average energy 1.25 MeV.
Half the photons are absorbed in the tumor, and half escape.
i. What energy is delivered to the tumor per second? [4 marks]
ii. What absorbed dose, in rad, is delivered per second? [2 marks]
iii. What equivalent dose, in rem, is delivered per second if the RBE for
these gamma rays is 0.70? [2 marks]
Page 6 of 7
iv. What exposure time is required for an equivalent dose of 200 rem? [2
marks]
B. A laser with power output of 2.0 mW at a wavelength of 400 nm is projected
onto a Calcium metal. The binding energy is 2.31 eV.
i. How many electrons per second are ejected? [6 marks]
ii. What power is carried away by the electrons? [4 marks]
C. A hypodermic needle of diameter 1.19 mm and length 50 mm is used to
withdraw blood from a patient? How long would it take for 500 ml of blood to be
taken? Assume a blood viscosity of 0.0027 Pa.s and a pressure in the vein of
1,900 Pa. [10 marks]
D. A person with lymphoma receives a dose of 35 gray in the form of gamma
radiation during a course of radiotherapy. Most of this dose is absorbed in 18
grams of cancerous lymphatic tissue.
i. How much energy is absorbed by the cancerous tissue? [2 marks]
ii. If this treatment consists of five 15-minute sessions per week over the
course of 5 weeks and just one percent of the gamma photons in the
gamma ray beam are absorbed, what is the power of the gamma ray
beam? [4 marks]
iii. If the gamma ray beam consists of just 0.5 percent of the photons
emitted by the gamma source, each of which has an energy of 0.03
MeV, what is the activity, in Curies, of the gamma ray source? [4 marks]
E. A water heater that is connected across the terminals of a 15.0 V power supply
is able to heat 250 ml of water from room temperature of 25°C to boiling point
in 45.0 secs. What is the resistance of the heater? The density of water is 1,000
kg/m2 and the specific heat capacity of water is 4,200 J/kg/°C. [10 marks]
Answer:
A i. E = 9.62 × 10⁻⁷ J/s
ii. The absorbed dose is 4.81 × 10⁻⁶ Gy
iii. The equivalent dose is 3.37 × 10⁻⁴ rem/s
iv. t = 593471.81 seconds
B. i. 4.025 × 10¹⁵/s
ii. 0.512 mW
C. 7218092.2 seconds
D. i. 6.3 × 10⁻¹ J
ii. 1.4 × 10⁻² W
iii. 1.57 × 10³ Curie
E. 0.129 Ω
Explanation:
The given parameters are;
Mass of tumor = 0.20 kg
Activity of Cobalt-60 = 2.60 × 10⁻⁴ Ci
Photon energy = 1.25 MeV
(i) The energy, E, delivered to the tumor is given by the relation;
[tex]E = \frac{1}{2}\left (Number \, of \, decay / seconds \right )\times \left (Energy \, of \, photon \right )[/tex]
[tex]E = \frac{1}{2}\left (2.6\times 10^{-4}Ci )\times \left (\frac{3.70\times 10^{10}decays/s}{1 Ci} \right )\times 1.25\times 10^{6}eV\times \frac{1.6\times 10^{-19}J}{1eV}[/tex]
E = 9.62 × 10⁻⁷ J/s
(ii) The equation for absorbed dose is given as follows;
Absorbed dose, D, in Grays Gy = (Energy Absorbed Joules J)/Mass kg
Therefore, absorbed dose = (9.62 × 10⁻⁷ J/s)/( kg) = 4.81 × 10⁻⁶ Gy
1 Gray = 100 rad
4.81 × 10⁻⁷ Gy = 100 × 4.81 × 10⁻⁶ = 4.81 × 10⁻⁴ rad/s
(iii) Equivalent dose, H, is given by the relation;
H = D × Radiation factor, [tex]w_R[/tex]
∴ H = 0.7 × 4.81 × 10⁻⁴ rad/s = 3.37 × 10⁻⁴ Sv = 3.37 × 10⁻⁴ rem/s
(iv) The exposure time required for an equivalent dose of 200 rem is given as follows;
[tex]\dot{H} = \dfrac{H}{t}[/tex]
Therefore;
[tex]t= \dfrac{200}{{3.37 \times 10^{-4}} } = 593471.81 \, s[/tex]
∴ t = 6.9 days
B. The number of electrons ejected is given by the relation;
[tex]N = \frac{P}{E} = \frac{P \times \lambda}{hc}[/tex]
[tex]N = \dfrac{2.0 \times 10^{-3} \times 400 \times 10^{-9}}{6.626 \times 10^{-34} \times 3 \times 10^8} = 4.025 \times 10^{15}/s[/tex]
(ii) The power carried by the electron
The energy carried away by the electrons is given by the relation;
[tex]KE_e = hv - \Phi[/tex]
[tex]KE_e = \frac{6.626 \times 10^{-34} \times 3 \times 10^8}{400 \times 10^{-9}} - 2.31 \times \frac{1.6 \times 10 ^{-19} }{1}[/tex]
[tex]KE_e = 4.9695 \times 10^{-19} - 3.696 \times 10 ^{-19} = 1.2735 \times 10^{-19} J[/tex]
Power, P[tex]_e[/tex], carried away by the electron = 4.025 × 10¹⁵ × 1.2735 × 10⁻¹⁹ = 0.512 mW
C. The given parameters are;
d = 1.19 mm, ∴ r = 1.19/2 = 0.595 × 10⁻³ m
l = 50 mm = 5 × 10⁻³ m
V = 500 ml = 5 × 10⁻⁴ m³
η = 0.0027 Pa
p = 1,900 Pa.
[tex]\dfrac{V}{t} = \dfrac{\pi }{8} \times \dfrac{P/l}{\eta } \times r^4[/tex]
[tex]t = \dfrac{8\times \eta\times V\times l }{\pi \times P \times r^4}[/tex]
[tex]t = \dfrac{8\times 0.0027 \times 5 \times 10^{-4} \times 5 \times 10^{-2} }{\pi \times 1900 \times (0.595 \times 10^{-4} )^4}[/tex]
t = 7218092.2 seconds
D) i. Energy absorbed is given by the relation;
E = m×D
Where:
D = 35 Gray = 35 J/kg
m = 18 g = 18 × 10⁻³ kg
∴ E = 35 × 18 × 10⁻³ = 6.3 × 10⁻¹ J
ii. Total time for treatment = 15 × 5 = 75 minutes
Energy absorbed = 6.3 × 10⁻¹ × 100 = 63 J
Power = Energy(in Joules)/Time (in seconds)
∴ Power = 63/(75×60) = 1.4 × 10⁻² W
iii. Whereby the power is provided by 0.5% of the photons emitted by the source, we have;
[tex]P_{source}= \frac{P_{beam}}{0.005} =\frac{0.0014}{0.005} =0.28 \, W[/tex]
1 MeV = 1.60218 × 10⁻¹³ J
0.03 MeV = 0.03 × 1.60218 × 10⁻¹³ J = 4.80654 × 10⁻¹⁵ J/photon
Therefore, the number of disintegration per second = 0.28 J/s ÷ 4.80654 × 10⁻¹⁵ J/photon = 5.83 × 10¹³ disintegrations per second
1 Curie = 3.7 × 10¹⁰ disintegrations per second
Hence, 5.83 × 10¹³ disintegrations per second = (5.83 × 10¹³)/(3.7 × 10¹⁰) Curie
= 1.57 × 10³ Curie
E. The parameters given are;
Density of water = 1000 kg/m³
Volume of water = 250 ml = 0.00025 m³
Initial temperature, T₁, = 25°C
Final temperature, T₂, = 100°C
Change in temperature, ΔT = 100 - 25 = 75°
Specific heat capacity of the water = 4200 J/kg/°C
Mass of water = Density × Volume = 1000 × 0.00025 = 0.25 kg
∴ Heat supplied = 4200 × 0.25 × 75 = 78,750 J
Time to heat the water = 45.0 sec
Therefore, power = Energy/time = 78750/45 = 1750 W
The formula for electrical power = I²R =VI = V²/R
Therefore, where V = 15.0 V, we have;
15²/R = 1750
R = 15²/1750 = 0.129 Ω.
The resistance of the heater = 0.129 Ω.
One of your classmates, Kevin, is trying to calculate the acceleration due to gravity at the top of Mt. Everest. Looking at an equation sheet, he sees that the acceleration due to gravity is g = G M r 2. For G, he plugs in the gravitational constant. For M, he plugs in the mass of the Earth. For r, Kevin plugs in the elevation (the height above sea level) of Mt. Everest. Will Kevin arrive at the right answer for g at the top of Mt Everest?
Answer:
no.
Explanation:
No because for M he put the mass of the earth instead of the mass of the object.
Kevin will not arrive at the right answer for g if he calculates the height from sea level, it must be from the center of the earth.
Gravitational acceleration:The force of gravity on an object of mass m is given by:
F = GMm/r²
where G is the gravitational constant
M is the mass of the earth
r is the distance from the center of the earth
This force is equal to the weight of the object given by:
mg = GMm/r²
so,
g = GM/r²
But here r is the distance of the object from the center of the earth not from the sea level.
So, Kevin will not arrive at the right answer for g if he calculates the height from sea level.
Learn more about gravitational acceleration:
https://brainly.com/question/13769294?referrer=searchResults
This is a measure of quantity of matter
Answer:
Mass
Explanation:
Mass is the measure of amount of matter contained within any substance and hence mass determines the weight. Unit of mass is kilogram as per ISI system of units.
Mass is measured through a balance. The more is the mass of an object, the more the balance tilts towards the object side.
Weight is equal to product of mass and the gravitational constant i.e 9.8m/s^2
a girl pushes an 18.15 kg wagon with a force of 3.63 N. what is the acceleration?
A. 0.06 m/s2
B. 5 m/s2
C. 0.2 m/s2
Answer:0.2 m/s^2
Explanation:
mass=18.15kg
Force=3.63N
Acceleration=force ➗ mass
Acceleration=3.63 ➗ 18.15
Acceleration=0.2 m/s^2
A girl pushes a wagon of mass 18.15 kg with a force of 3.63 N, so the acceleration of the wagon will be 0.2 m/s².
What is acceleration?In mechanics, acceleration is the measure of how rapidly an object's velocity changes over time. Accelerations as a vector quantity. An object's acceleration depends on the direction of the net force exerted on it.
A vector quantity, acceleration, is something that has both a magnitude and a direction. As a vector quantity, velocity is also. The ratio of the velocity vector change over a time interval to that interval is the definition of acceleration.
Mass, m =18.15 kg
Force, f = 3.63 N
Force = m × a
a= f / m
a =3.63 / 18.15
a = 0.2 m/s²
Hence, the acceleration of the wagon will be 0.2 m/s².
To get more information about acceleration :
https://brainly.com/question/12550364
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